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bonufazy [111]
3 years ago
10

How far did we jump to go from 17 to 20r=26-1717+r=26​

Mathematics
2 answers:
julsineya [31]3 years ago
7 0
Both are correct :)
BabaBlast [244]3 years ago
5 0

Answer:

Both are correct!

Step-by-step explanation:

if R = 26 - 17 = 9

now the next one

17 + R = 26

Same thing as R = 26 - 17 = 9

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Find the median for the following set of data:<br><br> 13,26,35,44,38,12,47,23,26
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\sf{\qquad\qquad\huge\underline{{\sf Answer}}}

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Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
Papessa [141]

Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

2) \text{P(at least one boy and one girl)}=\frac{3}{8}

3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

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4 years ago
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